Circles rolling on circles

Yutaka Nishiyama
Coin rolling

How many revolutions will the smaller coin make when rolling around the bigger one?




I figured the answer must be four revolutions. So imagine my surprise when I saw that the answer was given to be five!

I read the explanation of why this was indeed the correct answer, and although the reasoning seemed sound, it took some time before I could truly convince myself that my solution was in error. It’s an interesting problem, so I presented it to a number of people, most of whom immediately answered "four" as I did, and, like me, were difficult to convince otherwise; only a very few could intuitively see "five" as the correct answer.

Here’s a better way to think of this problem: rather than rolling along the larger circle, start by imagining the smaller circle as rolling on a line the same length as the circumference of the larger circle. In this case it is straightforward to think of the line as being $8\pi r$ units long, and so the smaller circle clearly must have made $8\pi r \div 2\pi r = 4$ rotations. Next, think of the circle as sliding along the line, without rolling, so that the point on the coin at which it touches the line remains the same. Now consider the difference between sliding along a straight line and doing the same along the circumference of a circle; if you slide units $8\pi r$ along a straight line, you arrive at your destination just as you started, without having ever changed orientation. But if you do the same along the circumference of a circle you will have made a full rotation at the time you return to your starting point. When rolling along the same circumference, therefore, you will have made the four rolling rotations plus the one sliding revolution, for a total of five!

Put differently: when the small circle rolls along the circumference of the larger circle, two kinds of movement simultaneously occur, revolution and rotation. The four movements one initially considers are the four revolutions, perhaps because these are readily seen. Rotation, on the other hand, is more difficult to grasp.

Coin rolling

It’s difficult to make progress on this kind of problem just by thinking about it, so it is important to verify the situation through experimentation. For example, you should try modelling this problem using two coins; if the problem followed the predictions of most people, then when using two same-sized coins the moving one would rotate $2\pi r \div 2\pi r = 1$ times, but as you will see it does so twice. For example, you might predict that rolling from the top of the fixed coin to its bottom would result in the rolling coin finishing upside-down, but in fact it will have unexpectedly performed a complete rotation by this point. I highly recommend trying this yourself.

If you find it difficult to grasp how things work on a circle, you might also want to imagine what would happen on a square. When a circle rolling along the outer periphery of a square encounters the first corner, it will have to rotate an "extra" 90° to continue along the next side. This will happen again at each corner, and since 90° x 4 = 360°, this accounts for an additional full revolution.

Each of the above explanations describes the circle's movement as a decomposition into rotation and revolution, but in reality no such decomposition is taking place. Just as a human's heart and lungs work simultaneously, rotation and revolution take place together. Separating revolution from rotation is helpful for understanding, but doing so does not provide a fundamental solution. Some say that the makeup of our brains does not allow for multitasking, but learning to simultaneously comprehend such phenomena would be of great value.

A similar problem appeared in Aha! Gotcha: Paradoxes to Puzzle and Delight by Martin Gardner and also in Scientific American in 1868. If you can think of an alternative proof or explanation for this problem, please post a comment or email us!

About the author


Yutaka Nishiyama is a professor at Osaka University of Economics, Japan. After studying mathematics at the University of Kyoto he went on to work for IBM Japan for 14 years. He is interested in the mathematics that occurs in daily life, and has written ten books about the subject. The most recent one is The Mysterious Number 6174: One of 30 mathematical topics in daily life, published by Gendai Sugakusha in July 2013 (ISBN978-4-7687-6174-8). You can visit his website here.


Think of the circle at the end of the 8xpixr line… It rotated four times. Now bend the line to make it a circle. The small circle has to rotate one more time to follow the line.

I prefer to think of this problem as a coin rotating around a spot; i.e. the 'big' circle has a radius of zero.

It's easy to see that the 'small' circle will rotate once, even though it's travelled a circumference on 0cm.

Then, if the 'big' circle has any additional circumference (in this case 4cm), the 'small' circle will rotate that once, then travel the additional circumference.

I hope this makes sense!

I arrived at your reasoning independently before looking at the comments.

Could one also consider that the edge of the circle are always curved away from the circle rolling on it, and because of this the circle must roll extra to meet this edge verses a straight line?


Ok I get that it's more complex than it first appears, but what about circles, with circles within them rotating. Do the maths, my head hurts.

This is also an interesting problem, like with a circle rotating on the exterior of another circle, the answer is not simply R/r, that is the ratio of the larger radius to the smaller radius, but is actually R/r - 1. By rolling along the interior of a circle, one revolution is lost. An intuitive way of imagining why this happens is by first approximating a circle with a regular polygon, say a hexagon. Imagine rolling a circle around the interior of the hexagon, while the circle rolls along each side of the hexagon, it does not get to roll over the complete length of each side, due to the convexity of the interior of the hexagon, namely, a circle will have completed rotation along one side when it is tangent to two sides of the hexagon, it is clear simply by drawing a picture that the circle does not need to be displaced the entire side length of each side of the hexagon for this to happen, and so some rotation is essentially *lost* because to this.

context not clear but should not "R/r - 1" be "R/r +1" (exterior rotation) ?

whereas "R/r -1" for interior rolling?

The CENTRE of the small coin needs to travel around a circle of radius 4+1=5cm. So the CENTRE needs to travel 10pi, as though it is traveling along a straight line of length 10pi while revolving. This gives the solution.

The centre of rotation of a point on the smaller circle is the centre of the smaller circle. Therefore for a full rotation the smaller circle will travel a distance of its circumference around the larger circle.
But the centre of rotation of the centre of the smaller circle is the centre of the larger circle. Hence it rotates around the circle by the larger radius as suggested.

4 or 5? David Van Leeuwen's animation by GeoGebra is a good explanation.
Yutaka Nishiyama

What your demo indicates is actually the location of the CENTER of small circle, which is causing illusion. How about the tip of the red arrow, which is tangential point of the two circles. Suppose you put a small ant at the tangential point (assuming ant not moving by itself). How many time this ant will be in contact with the large circle? Appearantly, the answer is 4, not 5. We are not interested in the movement of the center of small circle at all...

In the case of coins/circles of equal radius, the moving coin rotates once with respect to the static coin/circle but twice with respect to the observer.

The important point is that were it different, a driven planet gear of radius ‘r’, would rotate more than once when powered by a drive-cog of radius ‘r’. This would involve a creation of energy.

In the case of a coin/circle of radius ‘r’ moving along a line of length 2πr, the moving coin rotates once with respect to the static coin/circle and also once with respect to the observer.

In the animation at the black arrow is “with respect to the observer” and the red arrow is “with respect to the static circle.”

The mathematical formula Xπr / Yπr = X/Y remains true in all cases: anything else is an illusion.

The article reminds me of a surprising fact I learnt recently about the Moon's orbit round the Earth. It keeps the same face towards us all of the time (except for a relatively small effect known as libration) because it does one complete rotation for every revolution. I believe this synchronous orbit is explained as a gravitational effect called tidal locking.

This motion seems to be similar to what's happening when, as described in the article, you slide but don't roll a coin A round the edge of another B so that the same point on A's edge is always in contact with B. Coin A then does one full rotation, but not two.

It's almost as if gravity answers the reverse of the question posed in the article by explaining why we get one less rotation than we should.

Chris G

Looking at the picture of the £2 coins, imagine if the coin on top rotated one full time, but the bottom coin also rotated in the opposite direction - as if they were interlocking cogs, or a bit like the bottom coin is a treadmill on which the top coin turns.

In order for the top coin to rotate clockwise once, (i.e. 2 * pi * r), the bottom coin must also rotate anti-clockwise once. That is two full rotations. Now, if the bottom coin stays fixed (like a pavement and not a treadmill) as in the original statement of the problem, then is there some principle of physics that dictates the missing revolution of the bottom coin - because this bottom coin is now static - that says that revolution must come out somewhere? And that is why the top coin actually rotates twice? I'm thinking about a conservation law, but I do not know physics.

Michael B.

The problem here lies not in the maths used to get the answer or in incorrect thinking on the part of the problem solver - it lies in asking a vague question. There are several versions of this puzzle, all of them relying on vagueness to trick the reader. There may be several correct answers to the general question "How many rotations did the smaller circle make?":
1: the smaller circle only rotates around the centre of the larger circle once.
4: the smaller circle rotates around its own centre four times.
5: the smaller circle makes two types of rotation as above, totalling five rotations.
undetermined/zero/infinity: no start or end point for either circle has been defined, so the smaller circle rotates indefinitely, or does not rotate at all, because no rotation point has been defined either.

It can be argued that the standard definition for a rotation of a circle is a rotation around its own centre. A revolution, however, is a circular movement around an external point (external to the circle). So, in the above problem, the smaller circle makes 4 rotations and one revolution. (Of course, definitions of rotation and revolution can change depending on the context).

The question cannot be answered - fairly and mathematically - with a single answer, without first defining the rotation point and clarifying the definition of 'rotation' in this instance.

It does not help matters that, in the above example, 'revolution' is used in the diagram, while 'rotation' is used in the general text.

I detest these kinds of 'trick' questions, with the 'gotcha!' at the end. They make people feel foolish when they have only tried to think about the question and answer it with the information provided. If you fail to get the correct answer without being given all the information or when the question is badly or incorrectly phrased, then it is clearly not your fault.

Questions such as these do not support the use of mathematical skills (especially logic) in problem solving - they are essentially interpretative grammar tests, and the correct answer should be spotting the grammar or logic mistakes in the question.

Evidence for the above assertions lie in the fact that even people as intelligent as Yutaka Nishiyama can be duped by them. It is not a difficult problem, but a poorly written one.

It is exactly like the 'Deep Thought' computer in Hitchiker's Guide to the Galaxy providing the answer to the 'Ultimate Question' as 42. When confronted about providing such a flippant, simple answer, the computer states: "I think the problem, such as it was, was too broadly based. You never actually stated what the question was." All poorly stated questions in mathematics should be treated with similar contempt!

N.B.: Just to be clear, I am in no way criticising the esteemed Mr. Yutaka Nishiyama - I am only criticising the original question and all similarly badly written questions!

I am sorry but your argument is flawed. The question is very clear. It says the small circle is 'rolling' completely, which means it has to rotate around it's centre forever while moving, there's absolutely no sliding motion. Then the question asks, how many 'rotations' did the small circle make, not how many 'times' the small circle had to roll completely. If the question was the latter the answer wound have been 4. But given the change in orientation of the initial position that occurs in the small circle owing to moving on the big circle, we must consider a 5th rotation!

1) Radius:1 - 2(Pi)r = 2.68
2) Radius:4 - 2(Pi)r = 25.12

25.12/2.68 = 9.3 revolutions!?

Where I am I going wrong?

Your error is in solving your first equation: when r = 1 then 2(pi)r = 6.28 (to 2 d.p.), not 2.68.

You'll then find that 25.12 / 6.28 = 4, as expected.

I think it is possible to get a better picture by first thinking of a regular polygon, take a hexagon, for example. The hexagon has six corners, now if we initially started rotating a coin along the sides of the hexagon, once it reaches a corner, the coin rotates an additional 60 degrees ''around'' this corner, try and visualise this and hopefully it makes sense. Therefore, along with the rotation induced due to the coins motion around the sides of the hexagon, there is additional rotation along each of the six corners, and so one additional revolution is made once the coin reaches its starting position. Now if you think of a circle as the limit of regular polygons as number of sides tends to infinity, you can extend this logic to this problem to perhaps get a better intuitive sense as to why an additional revolution takes place.

Hi, okay, I'm really confused. What qualifies a revolution? How do we know that it has done one by the time it is at the bottom of the coin? Just because the coin is right side up at the bottom doesn't mean it has revolved completely (does it?). The way I see it is from the center coin's perspective; first it is being touched by the Queen's neck, then her face, then her hair, then the back of her head, and finally her neck again. But that leaves us at the start, causing me to think that it has really only performed one revolution. But obviously I'm wrong. Please can someone explain where I've gone wrong. Thanks.

here is another interpretation on rotation as found in this link please comment. thanks my email is

Thank you very much for your explanation. I was getting a little frustrated that no source I found went into detail explaining exactly why an extra "rotation" happened to be but your "sliding revolution" reasoning cleared the air beautifully.

'Sometimes, logic needs to take a back seat.'
-Me, probably.

My 16 year old son caught me with this last night & I marched straight into the incomplete response, but fairly quickly rebutted that the smaller circle does in fact revolve only [ratio] times in the radial space, but I conceded that a complete response would be to add that it revolves [ratio] + 1 times in the Cartesian space (after he clarified my assumption for me). I'll also concede that my retort was a complete reaction, but it now has me wondering if there are legitimate space-transformations to clarify the distinction?

Using the standard definition the one revolution is 360 degrees. Look carefully at the rotation of the coin about the other coin. Though it appears that the outer coin rotates twice as it goes around the center coin, it does not. It does one half a revolution from the top to the bottom, then completes another half from the bottom to the top. From the top coin as a reference, the right side of the coin is engaged in the first half of the rotation and the left side in the second half. A third way to see this is the contact point of the top coin is at the bottom of the coin, this point does not come back into contact with the center coin until it reaches the top position. The optical illusion comes from the orientation of the coin when it is at the bottom. Though the orientation is the same as when it is at the top, it is the top of the coin in contact the the center coin not the bottom point. The outer coin only does one 360 rotation as it rolls around the center coin.

Imagine a coin rolling around a pentagon, the coin will rotate in regular angular increments along each edge with an additional fifth of a rotation at each vertex. Now imagine increasing the number of edges. When the edges shorten and approach the vertices the Archimedean principle of exhaustion holds even when infinitely short edge and vertex are one. There is therefore an extra - you could say hidden - rotation. Of sorts.